Difference between revisions of "2024 IMO Problems/Problem 4"
Bobwang001 (talk | contribs) |
(→See Also) |
||
(6 intermediate revisions by 5 users not shown) | |||
Line 7: | Line 7: | ||
. | . | ||
==Video Solution== | ==Video Solution== | ||
− | https:// | + | https://youtu.be/WnZv3fdpFXo |
+ | |||
+ | ==Video Solution (AI solution)== | ||
+ | https://youtu.be/cjnJ6EXKWW4 | ||
+ | |||
− | ==Video Solution== | + | ==Video Solution(In Chinese)== |
− | https://youtu.be/ | + | https://youtu.be/QphkkutmY5M |
==Video Solution== | ==Video Solution== | ||
Line 25: | Line 29: | ||
https://youtu.be/iOp9mnmZyzU | https://youtu.be/iOp9mnmZyzU | ||
− | Comments: Although this is an IMO problem, the skills needed to solve this problem have all previously tested in AMC and its system math contests, such as HMMT. | + | Comments: Although this is an IMO problem, the skills needed to solve this problem have all previously tested in AMC and its system math contests, such as HMMT.~ also proved by Kislay Kai |
Evidence 1: 2020 Spring HMMT Geometry Round Problem 8 | Evidence 1: 2020 Spring HMMT Geometry Round Problem 8 | ||
Line 36: | Line 40: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution with discussion of a generalized case== | ||
+ | |||
+ | https://youtu.be/NJc79Ccg82E?si=J0YdHAz-46miJIO2 | ||
+ | |||
+ | ==Solution simple== | ||
+ | [[File:2024 AIME II 12 d.png|330px|right]] | ||
+ | Let point <math>Z</math> be <math>YZ||AB, XZ||AC, G \in AC, IG||BC, G' \in AB, IG' || BC.</math> | ||
+ | <math>F= AC \cap YZ, F' = AB \cap XZ.</math> | ||
+ | |||
+ | <math>AFZF'</math> is the parallelogram with equal heights, so <math>AFZF'</math> is rhomb <math>\implies</math> | ||
+ | <cmath>\angle CAZ = \angle AZY = \angle BCP, AI = IZ.</cmath> | ||
+ | <cmath>AL = LC, AI = IZ \implies IL || ZC \implies \angle LIG = \angle BCZ.</cmath> | ||
+ | <math>\angle AZY = \angle BCP = \angle YCP \implies</math> points <math>C,P, Y,</math> and <math>Z</math> are concyclic. | ||
+ | |||
+ | Therefore <math>\angle ZPY = \angle LIG.</math> | ||
+ | |||
+ | Similarly, <math>\angle ZPX = \angle KIG'. \blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2024|num-b=3|num-a=5}} |
Latest revision as of 04:29, 3 September 2024
Let be a triangle with
. Let the incentre and incircle of triangle
be
and
, respectively. Let
be the point on line
different from
such that the line
through
parallel to
is tangent to
. Similarly, let
be the point on line
different from
such that the line through
parallel to
is tangent to
. Let
intersect the circumcircle of
triangle
again at
. Let
and
be the midpoints of
and
, respectively.
Prove that
.
Contents
Video Solution
Video Solution (AI solution)
Video Solution(In Chinese)
Video Solution
Part 1: Derive tangent values and
with trig values of angles
,
,
Part 2: Derive tangent values and
with side lengths
,
,
, where
is the midpoint of
Part 3: Prove that and
.
Comments: Although this is an IMO problem, the skills needed to solve this problem have all previously tested in AMC and its system math contests, such as HMMT.~ also proved by Kislay Kai
Evidence 1: 2020 Spring HMMT Geometry Round Problem 8
I used the property that because point is on the angle bisector
,
is isosceles. This is a crucial step to analyze
. This technique was previously tested in this HMMT problem.
Evidence 2: 2022 AMC 12A Problem 25
The technique in this AMC problem can be easily and directly applied to this IMO problem to quickly determine the locations of points and
. If you read my solutions to both this AMC problem and this IMO problem, you will find that I simply took exactly the same approach to solve both.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution with discussion of a generalized case
https://youtu.be/NJc79Ccg82E?si=J0YdHAz-46miJIO2
Solution simple
Let point be
is the parallelogram with equal heights, so
is rhomb
points
and
are concyclic.
Therefore
Similarly,
vladimir.shelomovskii@gmail.com, vvsss
See Also
2024 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |